If the common tangents to the parabola, x ^ 2 = 4y and the circle,...

Let $y = mx + c$ be the common tangent.

Then ${c}^{2} = 4\left(1 + {m}^{2}\right) . . . ( 1 )$ (condition of tangency for circle)

Solving with ${x}^{2} = 4y$, we get ${x}^{2} = 4\left(mx + c\right)$

i.e. ${x}^{2} - 4mx - 4c = 0 . . . ( 1 )$

Being a tangent, $( 1 )$ must have equal roots.

$⇒ {\left(- 4m\right)}^{2} = 4\left(1\right)\left(- 4c\right) ⇒ {m}^{2} = - c . . . ( 2 )$

From \left(1\right) \text{ and } \left(2\right), {c}^{2} = 4 - 4c \text{ and } c < 0

$⇒ {c}^{2} + 4c - 4 = 0 ⇒ c = - 2\sqrt{2} - 2$ (As $c < 0$, so $c \ne 2\sqrt{2} - 2$).

So, both tangents have common $y$ intercept and thus intersect at $P\left(0, - 2 - \sqrt{2}\right)$.

Thus, distance of $P$ from origin is $2\left(\sqrt{2} + 1\right) units.</p>$